Question

If slit width is increased to .02 to 0.4 then percentage change fringe width will be.

• A. 20%
• B. 25%
• C. 60%
• D. 50%

Answer 1

Answer: (D)

50%

Answer 2

In a diffraction (or interference–diffraction) experiment the fringe (or diffraction‐envelope) width is given by

$$\beta=\frac{\lambda D}{a}$$

where

• $\lambda$ is the wavelength,

• $D$ is the screen distance, and

• $a$ is the slit width.

Thus, the fringe width is inversely proportional to the slit width. When the slit width is increased, the fringe width reduces according to

$$\beta \propto \frac{1}{a}.$$

Let the initial slit width be $a_1$ and final be $a_2$. Then

$$\frac{\beta_2}{\beta_1}=\frac{a_1}{a_2}.$$

If we take the numbers as intended to be (for a doubling of the slit width) i.e. from $0.2$ to $0.4$ (noting that sometimes a misprint occurs when a decimal point shifts) then

$$\frac{\beta_2}{\beta_1}=\frac{0.2}{0.4}=\frac{1}{2}.$$

Thus, the new fringe width is half the original fringe width. The percentage change in fringe width is

$$\% \text{ change} =\left(1-\frac{1}{2}\right)\times 100\% =50\%.$$

Since $\beta\propto\frac{1}{a}$, doubling the slit width (from 0.2 to 0.4) halves the fringe width. Thus, the fringe width reduces by 50%.